rcpp - Cohabitation of RcppArmadillo and RcppParallel

The following toy example for parallelFor works fine (f2 is the parallelized version of f1):

// [[Rcpp::depends(RcppParallel)]]

// [[Rcpp::depends(RcppArmadillo)]]

#include <RcppArmadillo.h>

#include <RcppParallel.h>

#include <iostream>

#define vector NumericVector

using namespace Rcpp;

using namespace RcppParallel;

// compute values i/i+1 for i = 0 to n-1

// [[Rcpp::export]]

vector f1(int n) {

vector x(n);

for(int i = 0; i < n; i++) x[i] = (double) i/ (i+1);

return x;

}

struct mytry : public Worker {

vector output;

mytry(vector out) : output(out) {}

void operator()(std::size_t begin, std::size_t end) {

for(int i = begin; i < end; i++) output[i] = (double) i/ (i+1);

}

};

// [[Rcpp::export]]

vector f2(int n) {

vector x(n);

mytry A(x);

parallelFor(0, n, A);

return x;

}

However, if I replace #define vector NumericVector by #define vector arma::vec this doesn’t work any more. The codes compiles and run, f1 is ok, but the vector returned by f2 just contains uninitialized values.

Many thanks in advance for any clarification.

网友答案:

The problem here -- your class should be taking the vector by reference, rather than by value.

This is because, when using RcppParallel, you typically pre-allocate memory for an object somewhere, then fill that object -- so the parallel workers should be taking a reference to that object you want to fill.

Note that this works (perhaps surprisingly) for Rcpp vectors because they already are just 'proxy' objects -- just objects encapsulating a pointer to data. When you pass an Rcpp vector by value, you copy the pointer (not the underlying data!) plus some extra vector bits (e.g. the length of the vector) -- so the 'copy' retains a reference to the same data structure.

When you use a more 'classical' vector, e.g. the arma::vec or std::vector, when passing that by value to the worker you really are copying a whole new vector to the class, then filling that (temporary, copied) vector -- so the original vector never actually gets filled.